What are the normality and molarity of a solution prepared by dissolving 8.050g Ba(OH)2•8H2O in 1500ml of solution?
M = mols/L = 8.050/molar mass Ba(OH)2 /1.500 L = ?
Technically, one cannot determine the equivalent weight without know the reaction/titration; however, usually when no reaction is given we count the number of H or OH in the molecule. In the case of Ba(OH)2 it would be eq. wt. = molecular weight/2, then
N = # equivalents/L so that would be
N = (8.050 g/molar mass/2)/1.500
To determine the normality and molarity of a solution prepared by dissolving 8.050g of Ba(OH)2•8H2O in 1500ml of solution, we need to follow these steps:
Step 1: Calculate the molar mass of Ba(OH)2•8H2O
To find the molar mass, we add up the atomic masses of all the atoms in the compound:
Ba: 1 x 137.33 g/mol = 137.33 g/mol
O: 2 x 16.00 g/mol = 32.00 g/mol
H: 2 x 1.01 g/mol = 2.02 g/mol
H2O: 8 x 18.02 g/mol = 144.16 g/mol
Total molar mass = 137.33 g/mol + 32.00 g/mol + 2.02 g/mol + 144.16 g/mol = 315.51 g/mol
Step 2: Convert the mass of Ba(OH)2•8H2O to moles
To convert grams to moles, we divide the mass by the molar mass:
Moles = Mass / Molar mass
Moles = 8.050 g / 315.51 g/mol ≈ 0.0255 moles
Step 3: Calculate the volume of the solution in liters
To convert milliliters (ml) to liters (L), we need to divide by 1000:
Volume (L) = 1500 ml / 1000 = 1.5 L
Step 4: Calculate the normality (N) of the solution
Normality is defined as the number of equivalents of solute per liter of solution. Since Ba(OH)2 is a strong base that dissociates completely in water and each formula unit of Ba(OH)2 releases 2 equivalents of OH-, the normality can be calculated as:
Normality (N) = moles of solute / Volume of solution (L)
Normality (N) = 0.0255 moles / 1.5 L ≈ 0.017 M
Step 5: Calculate the molarity (M) of the solution
Molarity is defined as moles of solute per liter of solution. In this case, since Ba(OH)2•8H2O dissociates into three ions (Ba2+, 2OH-) in water, we need to multiply the moles of Ba(OH)2 by 3:
Molarity (M) = (moles of solute x number of ions) / Volume of solution (L)
Molarity (M) = (0.0255 moles x 3) / 1.5 L ≈ 0.051 M
So, the normality of the solution is approximately 0.017 N, and the molarity of the solution is approximately 0.051 M.
To determine the normality (N) and molarity (M) of the solution prepared, you need to follow a step-by-step approach:
Step 1: Calculate the number of moles of Ba(OH)2·8H2O
To do this, you need to divide the given mass (8.050g) by the molar mass of Ba(OH)2·8H2O.
The molar mass of Ba(OH)2·8H2O can be calculated by adding the atomic masses of its constituent elements:
Ba = 137.327 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
So, molar mass of Ba(OH)2·8H2O = (1 x Ba) + (2 x O) + (16 x H) + (8 x H2O)
= (1 x 137.327) + (2 x 16.00) + (16 x 1.01) + (8 x 18.015)
= 171.329 g/mol
Now, divide the given mass (8.050g) by the molar mass to find the number of moles:
Moles of Ba(OH)2·8H2O = 8.050g / 171.329 g/mol
Step 2: Calculate the volume in liters
The given volume is expressed in milliliters (1500 mL). Convert it to liters by dividing by 1000:
Volume (V) = 1500 mL / 1000 mL/L
Step 3: Calculate the normality (N) and molarity (M)
The normality is defined as the number of equivalents of solute per liter of solution. As Ba(OH)2 is a strong base that dissociates completely in water, the normality is equal to the number of moles divided by the volume in liters:
Normality (N) = Moles of Ba(OH)2·8H2O / Volume (V)
Similarly, the molarity is defined as the number of moles of solute per liter of solution:
Molarity (M) = Moles of Ba(OH)2·8H2O / Volume (V)
Finally, substitute the calculated values into the equations, and solve for normality and molarity.